Answer
$L(x)=\frac{1}{12}x+\frac{4}{3}$
Work Step by Step
Step 1. $f(x)=\sqrt[3] x=x^{1/3}$, $f'(x)=\frac{1}{3}x^{-2/3}$,
Step 2. Choose $x=8$, $f(8)=2$, $f'(8)=\frac{1}{3}8^{-2/3}=\frac{1}{12}$,
Step 3. The linearization at $x=8$ is $L(x)=f(8)+f'(8)(x-8)=2+\frac{1}{12}(x-8)=\frac{1}{12}x+\frac{4}{3}$
